The orbital speed of Earth about the Sun is $3.0 \times 10^{4} \mathrm{m} / \mathrm{s}\( and its distance from the Sun is \)1.5 \times 10^{11} \mathrm{m}$. The mass of Earth is approximately \(6.0 \times 10^{24} \mathrm{kg}\) and that of the Sun is \(2.0 \times 10^{30} \mathrm{kg} .\) What is the magnitude of the force exerted by the Sun on Earth? \([\) Hint: Two different methods are possible. Try both.\(]\)

We can use Newton's law of gravitation to determine the force between Earth and the Sun. The formula for gravitational force F between two objects with masses m1 and m2, separated by distance r, is: F = G * (m1 * m2) / r^2 where G is the gravitational constant and is equal to 6.674 * 10^{-11} N * m^2/kg^2. Given in the exercise, the mass of Earth (m1) is 6.0 * 10^{24} kg, the mass of the Sun (m2) is 2.0 * 10^{30} kg, and their distance (r) is 1.5 * 10^{11} m. Plugging these values into the formula, we get: F = (6.674 * 10^{-11}) * [(6.0 * 10^{24}) * (2.0 * 10^{30})] / (1.5 * 10^{11})^2 Calculating the force, we get: F ≈ 3.54 * 10^{22} N

Centripetal force (Fc) is the force acting on an object moving in a circular path, directed towards the center of the circle, and can be computed using the following formula: Fc = (m * v^2) / r Where m is the mass of the object (in this case, Earth), v is its orbital speed, and r is the distance between the object and the center of the circle (i.e., distance between Earth and the Sun). Given in the exercise, Earth's mass (m) is 6.0 * 10^{24} kg, the orbital speed (v) is 3.0 * 10^{4} m/s, and the distance (r) is 1.5 * 10^{11} m. Plugging these values into the formula, we get: Fc = (6.0 * 10^{24}) * (3.0 * 10^{4})^2 / (1.5 * 10^{11}) Calculating the centripetal force, we obtain: Fc ≈ 3.54 * 10^{22} N As we can see, both methods give us the same result: The magnitude of the force exerted by the Sun on Earth is approximately 3.54 * 10^{22} N.